This question was motivated by an answer to this question of Dominic van der Zypen.

It relates to the following classical theorem of Sierpiński.

Theorem (Sierpiński, 1921). For any countable partition of the unit interval $[0,1]$ into closed subsets exactly one set of the partition is non-empty.

Motivated by this Sierpiński Theorem we can ask about the smallest infinite cardinality $\acute{\mathfrak n}$ of a partition of the unit interval into closed non-empty subsets. It is clear that $\acute{\mathfrak n}\le\mathfrak c$. The Sierpinski Theorem guarantees that $\omega_1\le\acute{\mathfrak n}$. So, $\acute{\mathfrak n}$ is a typical small uncountable cardinal living in the segment $[\omega_1,\mathfrak c]$.

Problem 1. Is $\acute{\mathfrak n}$ equal to some other known small uncountable cardinal?

According to Theorem 4 of Miller, the strict inequality $\acute{\mathfrak m}<\mathfrak c$ is consistent. So, $\acute{\mathfrak m}$ is a non-trivial small uncountable cardinal.

Problem 3. Is it consistent that $\acute{\mathfrak n}<\acute{\mathfrak m}$?

Problem 4. Is $\acute{\mathfrak m}$ equal to some known small uncountable cardinal?

Added after analyzing comments to these problems: As was observed by @Ashutosh, the answer to Problem 2 is affirmative. In his paper Miller writes that this was done by Both (1968, unpublished) and Weiss (1972, unpublished). The MA equality $\acute{\mathfrak n}=\mathfrak c$ can be also derived from the ZFC inequality $$\mathfrak d\le\acute{\mathfrak n},$$ which can be proved as follows: given a partition $\mathcal P$ of $[0,1]$ into pairwise disjoint closed sets with $|\mathcal P|=\acute{\mathfrak n}$, we can choose a countable subfamily $\mathcal P'\subset\mathcal P$ such that the space $X=[0,1]\setminus\bigcup\mathcal P'$ is nowhere locally compact and hence is homeomorphic to $\omega^\omega$. Then $\mathcal P\setminus\mathcal P'$ is a cover of $X\cong\omega^\omega$ by compact subsets, which implies that $\acute{\mathfrak n}=|\mathcal P\setminus\mathcal P|'\ge\mathfrak d$ by the definition of the cardinal $\mathfrak d$.

Miller proved the consistency of the strict inequality $\acute{n}<\mathfrak c$. Looking at the diagram of small uncountable cardinals in Vaughan, I found only three small uncountable cardinals above $\mathfrak d$: $\mathfrak i$, $cof(\mathcal M)$ and $cof(\mathcal L)$.

Problem 5. Is $\acute{\mathfrak n}$ equal to one of the cardinals $\mathfrak d$, $\mathfrak i$, $cof(\mathcal M)$ or $cof(\mathcal L)$ in ZFC?

Summing up the progress made sofar. The cardinals $\acute{\mathfrak n}$ and $\acute{\mathfrak m}$ satisfy the following ZFC-inequalities:

1 Answer
1

A lot of very good observations have already been put into the comments. I'll add one more observation that's too long for a comment:

It is consistent that $\acute{\mathfrak{m}} < \mathrm{non}(\mathcal L)$.

The basic idea is to start with a model of MA + $\neg$CH (where $\mathrm{non}(\mathcal L)$ is already big), and then to force over this model to make $\acute{\mathfrak{m}}$ smaller while leaving $\mathrm{non}(\mathcal L)$ large. The proof uses two facts:

Fact 1: There is a $\sigma$-centered notion of forcing $P$ that does not change the value of $\mathfrak{c}$ and that forces $\acute{\mathfrak{m}} = \aleph_1$.

Fact 2: Suppose $V$ is a model of MA, and $P$ is a $\sigma$-centered notion of forcing (in $V$). Then $P$ does not lower the value of $\mathrm{non}(\mathcal L)$. In particular, if $V \not\models$ CH and if $P$ does not change the value of $\mathfrak{c}$, then $\mathrm{non}(\mathcal L) = \mathfrak{c} > \aleph_1$ in the extension.

My claim follows easily from these two facts. To get a model of $\acute{\mathfrak{m}} < \mathrm{non}(\mathcal L)$, begin with a model of MA + $\mathfrak{c} > \aleph_1$ and force with the notion of forcing $P$ described in Fact 1. The resulting model has $\acute{\mathfrak{m}} = \aleph_1$ (because $P$ makes this true) and $\mathrm{non}(\mathcal L) = \mathfrak{c} > \aleph_1$ (by the "in particular" part of Fact 2). QED

Fact 2 is possibly "well known" but I don't know the standard reference. It was first explained to me by Andreas Blass, who exposits it nicely in the proof of Corollary 49 in this paper.

For Fact 1, the notion of forcing from Theorem 4 in this paper of Arnie Miller does the job. This forcing -- let us call it $P$ -- is $\sigma$-centered and does not change the value of $\mathfrak c$.$^{(*)}$ $P$ is designed to add a partition of $2^\omega$ into $\aleph_1$ closed sets, and it is easy to show (using a genericity argument) that each set in this partition has measure $0$; Miller even points this out in a comment after the proof of his Theorem 4.

There may be a small issue here about partitioning $2^\omega$ instead of $[0,1]$, but we can get around it. Given our partition of $2^\omega$ into $\aleph_1$ closed measure-zero sets, first observe that none of them has interior in $2^\omega$ (because then it would fail to have measure $0$). Thus there is a countable, dense $D \subset 2^\omega$ that contains no more than one point of any member of our partition. Recall that if $C$ is a closed measure-zero subset of $2^\omega$, then $C$ minus one point is homeomorphic to a countable disjoint union of Cantor sets, so it can be partitioned into countably many closed measure-zero sets. Thus, we may modify our partition by adding in the sets of the form $\{d\}$ with $d \in D$, and then dividing some other sets into countably many pieces. In this way we obtain a partition of $2^\omega$ into $\aleph_1$ closed measure-zero sets, including all sets of the form $\{d\}$ for $d \in D$. Once we have done this, we observe that there is a measure-preserving homeomorphism from $2^\omega \setminus D$ onto $[0,1] \setminus \mathbb Q$. When we push our partition through this homeomorphism, we obtain a partition of $[0,1]$ into $\aleph_1$ closed measure-zero sets.

$(*)$ Neither of these facts is stated explicitly in the linked paper, but neither is difficult to prove either. (Using Miller's notation from the linked paper:) Each forcing of the form $P(X)$ is $\sigma$-centered, because if two conditions agree on the part that asserts sentences of the form "$[s] \cap C_n = \emptyset$'' then they are compatible (take the union of the other part). It is clear that no $P(X)$ increases $\mathfrak c$, because it is too small ($|P(X)| \leq \mathfrak c$ and it has the c.c.c., so there are only $\mathfrak c$ nice names for reals). Thus $P$, which is a length-$\omega_1$, finite support iteration of forcings of the form $P(X)$, also is $\sigma$-centered and does not increase $\mathfrak c$.

$\begingroup$Thank you for this comment. Maybe it yields more, namely, that $\acute{\mathfrak m}<non(\mathcal L)$ is consistent? By the way, writing about the Baire space, maybe you had in mind $[0,1]\setminus\mathbb Q$ (not $\mathbb Q\cap[0,1]$)? Usually $\mathbb Q$ denotes the set of rational numbers.$\endgroup$
– Taras BanakhNov 13 '17 at 19:39

$\begingroup$Oops! Yes, I did mean $[0,1] \setminus \mathbb Q$, and I fixed the typo now.$\endgroup$
– Will BrianNov 13 '17 at 20:00

$\begingroup$As for $\acute{\mathfrak{m}}$, it would follow from this argument that $\acute{\mathfrak{m}} < \mathrm{non}(\mathcal L)$ if you could show that the forcing $P$ mentioned in my answer (or some other $\sigma$-centered forcing) adds an $\aleph_1$-sized partition of $[0,1]$ into closed measure-zero sets. I did consider this, but don't know whether it's true. The partition it adds for $2^\omega$ is a partition into closed measure-zero sets, but it's not obvious (to me, anyway) how to get from there to a partition on $[0,1]$. I will think about it some more, though.$\endgroup$
– Will BrianNov 13 '17 at 20:03

$\begingroup$@TarasBanakh: OK, I think I can see how to get $\acute{\mathfrak{m}} < \mathrm{non}(\mathcal L)$ now. I have to go teach now, but will edit my answer when I have the time (tomorrow at the latest).$\endgroup$
– Will BrianNov 13 '17 at 20:20

1

$\begingroup$@TarasBanakh: $\mathrm{non}(\mathcal M)$ is $\aleph_1$ in this extension because Cohen reals are added at every stage of the iteration (a set $\{r_\alpha : \alpha \in \omega_1\}$ of reals where $r_\alpha$ is Cohen over $V^{P_\alpha}$ will be non-meager in the extension). I'm not sure about $\mathfrak{i}$, but I'll think about it.$\endgroup$
– Will BrianNov 13 '17 at 21:56